原子物理习题库及解答
第一章
由能量、动量守恒
(这样得出的是电子所能得到的最大动量,严格求解应用矢量式子)
Δp θ
得碰撞后电子的速度
p
故
由
1-2 (1)
(2)
1-3 Au核:
Li核:
1-4 (1)
(2)
1-5
1-6
时,
时,
1-7 由
,得
由
,得
1-8(1)设碰撞前m1的速度为v1,动量为p1。
碰撞后m1的动量为
,m2的动量为
由动量、能量守恒可得:
其中
,将它代入上两式可得:
它们之间的矢量关系可用下图
示,其中圆心C为质心,
表示质心系里m1碰撞后的速度。
当
时,A点在圆上
时,A点在圆上
时,A点在圆外
由图可知,
(2)因
(请参阅朗道的力学)
1-9 对Au核:
对Ag核:
由
可求得
(其中
;
)
1-10
EMBED Equation.3
(1)
(2)
(3)
第二章
2-1 (1)
(Å)
(2)
(Å)
2-2 利用
H原子:
(Å),
(Å),
He+离子:
(Å),
(Å),
Li++离子:
(Å),
(Å),
(2) H原子:
He+离子:
Li++离子:
(3) H原子:
(Å)
He+离子:
(Å)
Li++离子:
(Å)
2-3
2-4
由能量、动量守恒可得质子的阈能:
2-5 (1)
现
故
(米3)
(2)室温下氢原子
2-6 只观察到赖曼系的头四条谱线 1216 Å,1026 Å,973 Å,950 Å
2-7
故
2-8 利用
2-9 利用折合质量
(1)
(2)V电离=13.6/2=6.8(V)
(3)
(Å)
2-10
(1)
(Å)
(2)
(3)
(Å)
2-11
, 将
代入
2-12 (1)
(2)反冲能
2-13 利用选择定则
,共有6条。
2-14 (1)
上两式相加得,
(2)
第三章
3-1
3-2
3-3
3-4
= 124 (T/m) (gJ = 2, mJ = ±1/2)
3-5
3-6
,且
,故分裂成四条。
3-7
又
是Li++.
3-8
又
(略)
能级图:3S1分裂成三条,g = 2
3p0不分裂
故不是正常塞曼效应。
3-11 参照
图15.7
由此可计算得
3-12 (1)
(2)
而
3-13 (1)
(2)
(3)3S态的能级分裂:
3P态的能级分裂:
能级图见p115图书馆5.11。
3-14
(Å)
谱线分裂为三条:
(Å)
(Å)
(Å)
第四章
4-1
4-2
4-3
4-4
4-5 价电子数为偶数的氦,Be, Mg, Ca,可能出现正常塞曼效应。
S可能为0。
,先算L-S耦合
EMBED Equation.3
j-j耦合
共18种态,且出现相同J的次数也和L-S耦合相同。
4-7 (1)np4形成的电子态与np2相同
如考虑泡里原理只有
,其中
能量最低。
(2)np5形成的电子态同np1相同,故只有
,其中
态能量最低。
(3)同上题的LS耦合,由于非同科电子,故有
,其中
能量最低。
4-8 (2S,3P)所形成的原子态为
L S = 0 S = 1
根据跃迁的选择定则
, 共可产生10条光谱线:
(若该电子被激发到2P态,则只发一条光谱)
4-9
故J = 0,
必定有
的基态。
4-10
2 1 0 -1 -2
4 3 2 1 0 2
3 2 1 0 -1 1
2 1 0 -1 -2 0
1 0 -1 -2 -3 -1
0 -1 -2 -3 -4 -2
S = 0时,ML = 4 3 2 1 0 –1 –2 –3 –4 L = 4
= 2 1 0 –1 –2 L = 2
= 0 L = 0
S = 1时, 对角线上值不能取
ML = 3 2 1 0 –1 –2 –3 L = 3
= 1 0 –1 L = 1
有
其中
最低。
4-11
基态氦原子是
,故不分裂,只有1束。
而硼原子的基态是
,分裂为2束。
4-12
的电子组态为
故基态为:
的电子组态为:
(倒转次序)
故基态为:
的电子组态为:
,倒转次序,故基态为
第五章
5-1
5-2 利用
解得
5-3
(应是电离一个L电子所做的功)
5-4
的末态
,而
的末态
谱线强度同末态数成正比,故
比
强2倍。
5-5 (1)
(2)
的波长
(Å)
激发L系所需的最小能量为
5-6 利用
(Å)
5-7 利用
散射光子能量最小。
电子的动能
5-8
, 联立可得
5-9
, 可得
解得:
5-10
故无论
多大,
不能产生正负电子偶。
5-11 采用质心系,反应后动量都为0,故只要考虑能量守恒。
(反应后电子在质心系动量为0。
)
故有
,不可能产生光电效应。
5-12 能量守恒
动量守恒
故不能在真空中发生光子
电子对过程。
5-13 (1)先考虑
:
再同
耦合:
满壳层缺2个,倒转次序,故基态为
(2)
故
(3)
,
5-14
(克/厘米2)
第六章
6-1
当
时,
(Å)
时,
(Å)
时,
(Å)
6-2 (1)
(2)
6-3 (1)
(2)
(Å)
6-4
(Å)
6-5 由
得
其中
6-6 (1)
(2)
故
6-7
6-8
6-9 (1)
(2)
(3)
6-10 (1)
(2)
(3)
6-11 设势阱边界为[0,a],则
而
按经典理论,
,粒子在空间出现的几率相同为
故
与
时的量子结论一致。
6-12
由
及二阶导数小于0得
(第一玻尔半径)
由
及二阶导数小于0得
6-13
(略)
立方程: Ⅱ:
Ⅲ:
方程解为: Ⅰ:
Ⅱ:
Ⅲ:
利用波函数
条件:
得B=0
---------(1)
---------(2)
得:
或
由图解法,
可判断存在束缚态的条件为:
即
6-16 由透射几率
估计,由于分子都是宏观量
第七章
7-1 根据原子的质量求
7-2
EMBED Equation.3
故
(年)
7-3 1克碳中碳14的含量为
(克)
故
(个)/克
故
而
由
(年)
7-4 由
天,
/天
7-5
故
7-6
故
7-7
变化时放出的能量为
K电子的结合能
故中微子的能量为
7-8 (1)
反应能
(2)
反应能
7-9
入射质子的能量
又
而
7-10 参阅史包尔斯基采用原子质量的精确表达式:
由
得,
由相当
衰变最稳定的那些原子的Z、A代入上式可求得
故
7-11 (1)设碳原子的质量为
,碰撞后的速度为
,能量为
,由
能量守恒
动量守恒
可得
(A为碳原子的质量数)
因此经一次碰撞后的中子能量为
故N次碰撞后的能量为
(2)
(次)
(1)质子的结合能
的激发能级
(2)由
可得各能级的平均寿命为:
7-13
1千克氚释放的能量为
100万千瓦电站一年中释放的能量为
焦耳/秒×3.15×107秒
故一年中消耗的氚为
(千克)
折合煤为
(千克)
(吨)
7-14
7-15
(月-1)即要有
个质子
才能每月观察到一次衰变,它相当于
(克)
所需的水为
(克)
(吨)
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